VbvAstE-001

It can be found for the known density distribution inside a planet γ(r). Furthermore,
accounting to the thermodynamic equation for chemical potential
dχ = m′
dp (12.17)
γ
(where m ′ is the ion mass) from the equation of state Eq.(12.4), the chemical potential
is
(
)
χ = αm ′ (k + 1)γ 1/k − γ1/k − k 2 γ
(12.18)
1 − k
and the density of the internal energy of the core is
ε in = χγ ( 3
m − p = ′ αn 2 γ5/3 n + 3γn 2/3 − 9 )
2 γn . (12.19)
Doing analogous calculations for the mantle, we obtain
ε im = α m
( γ
2
m (r)
γ 2 0
+ γm(r)
γ 0
− γm(r) ln γm(r) )
− 2
γ 0 γ 0
The electric energy exists only inside the core and its density is
. (12.20)
E 2 (r)
8π = 2π 9 Gγ2 nr 2 . (12.21)
Since the thermal energy is neglected, to calculate the full energy of the planet, it is
necessary to integrate Eqs.(12.19), (12.20), and (12.22) over the volume of the planet
and sum them and Eq.(12.16). To do this, we need to determine the values of constants
composing these equations.
12.5 The density distribution inside the Earth
The mass M and radius R of the Earth are known. Therefore, we know the average
density of the Earth < γ > ∼ = 5.5g/cm 3 . On the basis of the geophysical data, we
accept that the density of matter and bulk module on the surface of the mantle is
γ 0
∼ = 3.2g/cm 3 and B = 1.3 · 10 12 dyn/cm 2 . These values are characteristic for basalts
[24]. Based on the above said we determine R 0 and the parameter α m. We can found
the value of the parameter α m as we know the values of γ 0 and < γ > and therefore
we can find the ratio
( ) 1/3
R γ0
=
= 0.835. (12.22)
R 0 < γ >
Next from all possible solutions we choose the one that actually meets the condition
(12.22). In fact this procedure is reduced to choosing the parameter m ′ , i.e. the
ion mass related to each free electron in electron-ion plasma of the core. The total
energy (related to GM/R 0) is plotted as a function of the parameter m ′ in fig.(12.1).
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